$ABC$ is a triangle. Locate a point in the interior of $\Delta ABC$ which is equidistant from all the vertices of $\Delta ABC$.
(N/A) Let us consider a $\Delta ABC$.
Draw line $l$ as the perpendicular bisector of side $AB$.
Draw line $m$ as the perpendicular bisector of side $BC$.
Let the two perpendicular bisectors $l$ and $m$ intersect at point $O$.
Since any point on the perpendicular bisector of a segment is equidistant from its endpoints,point $O$ lies on the perpendicular bisector of $AB$ (so $OA = OB$) and on the perpendicular bisector of $BC$ (so $OB = OC$).
Therefore,$OA = OB = OC$,which means point $O$ is the required point equidistant from all vertices $A, B,$ and $C$. This point $O$ is known as the circumcenter of the triangle.
Draw line $l$ as the perpendicular bisector of side $AB$.
Draw line $m$ as the perpendicular bisector of side $BC$.
Let the two perpendicular bisectors $l$ and $m$ intersect at point $O$.
Since any point on the perpendicular bisector of a segment is equidistant from its endpoints,point $O$ lies on the perpendicular bisector of $AB$ (so $OA = OB$) and on the perpendicular bisector of $BC$ (so $OB = OC$).
Therefore,$OA = OB = OC$,which means point $O$ is the required point equidistant from all vertices $A, B,$ and $C$. This point $O$ is known as the circumcenter of the triangle.
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